Question: Let $y=\sqrt{x}\cos(x)$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-2\sqrt{x}\cos(x)-\sqrt{x}\sin(x)$ (Choice B) B $-\dfrac{1}{\sqrt{x}}+\sin(x)$ (Choice C) C $-\dfrac{\sin(x)}{\sqrt{x}}$ (Choice D) D $\dfrac{\cos(x)}{2\sqrt{x}}-\sqrt{x}\sin(x)$
$\sqrt{x}\cos(x)$ is the product of two, more basic, expressions: $\sqrt{x}$ and $\cos(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}(\sqrt{x}\cos(x)) \\\\ &=\dfrac{d}{dx}(\sqrt{x})\cos(x)+\sqrt{x}\dfrac{d}{dx}(\cos(x))&&\gray{\text{The product rule}} \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{1}{2}}}\right)\cos(x)+\sqrt{x}\dfrac{d}{dx}(\cos(x))&&\gray{\text{Write }\sqrt{x}\text{ as a power}} \\\\ &=\dfrac12x^{^{-\frac{1}{2}}}\cdot \cos(x)+\sqrt{x}\cdot (-\sin(x))&&\gray{\text{Differentiate }x^{^{\frac{1}{2}}}\text{ and }\cos(x)} \\\\ &=\dfrac{\cos(x)}{2\sqrt{x}}-\sqrt{x}\sin(x)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{\cos(x)}{2\sqrt{x}}-\sqrt{x}\sin(x)$